創建GROUP BY查詢，以顯示最新的排

Question

所以我的表是：
user_msgs：http://sqlfiddle.com/#!9/7d6a9
token_msgs：http://sqlfiddle.com/#!9/3ac0f

只有這些用戶4所列。當用戶將消息發送給其他用戶，查詢檢查是否存在已通過檢查token_msgs表的from_id和to_id開始的2個用戶之間的通信，如果沒有令牌存在，創建token和使用，在user_msgs表。所以令牌是這兩個表中唯一的字段。

現在，我想列出與之user1已經開始對話的用戶。因此，如果from_id或to_id包括1這些對話應該列出。

有在user_msgs表相同的用戶會話多行。

我想我需要使用group_concat，但不能確定。我試圖建立查詢，以做同樣展示了最新的談話頂部，因此ORDER BY time DESC：

SELECT * FROM (SELECT * FROM user_msgs ORDER BY time DESC) as temp_messages GROUP BY token 

請構建查詢幫助。

謝謝。

CREATE TABLE `token_msgs` (
    `id` int(11) NOT NULL, 
    `from_id` int(100) NOT NULL, 
    `to_id` int(100) NOT NULL, 
    `token` varchar(50) NOT NULL 
) ENGINE=MyISAM DEFAULT CHARSET=latin1; 

-- 
-- Dumping data for table `token_msgs` 
-- 

INSERT INTO `token_msgs` (`id`, `from_id`, `to_id`, `token`) VALUES 
(1, 1, 2, '1omcda84om2'), 
(2, 1, 3, '1omd0666om3'), 
(3, 4, 1, '4om6713bom1'), 
(4, 3, 4, '3om0e1abom4'); 


--- 


CREATE TABLE `user_msgs` (
    `id` int(11) NOT NULL, 
    `token` varchar(50) NOT NULL, 
    `from_id` int(50) NOT NULL, 
    `to_id` int(50) NOT NULL, 
    `message` text NOT NULL, 
    `time` datetime NOT NULL 
) ENGINE=MyISAM DEFAULT CHARSET=latin1; 

-- 
-- Dumping data for table `user_msgs` 
-- 

INSERT INTO `user_msgs` (`id`, `token`, `from_id`, `to_id`, `message`, `time`) VALUES 
(1, '1omcda84om2', 1, 2, '1 => 2\r\nCan I have your picture so I can show Santa what I want for Christmas?', '2016-08-14 22:50:34'), 
(2, '1omcda84om2', 2, 1, 'Makeup tip: You\'re not in the circus.\r\n2=>1', '2016-08-14 22:51:26'), 
(3, '1omd0666om3', 1, 3, 'Behind every fat woman there is a beautiful woman. No seriously, your in the way. 1=>3', '2016-08-14 22:52:08'), 
(4, '1omd0666om3', 3, 1, 'Me: Siri, why am I alone? Siri: *opens front facing camera*', '2016-08-14 22:53:24'), 
(5, '1omcda84om2', 1, 2, 'I know milk does a body good, but damn girl, how much have you been drinking? 1 => 2', '2016-08-14 22:54:36'), 
(6, '4om6713bom1', 4, 1, 'Hi, Im interested in your profile. Please send your contact number and I will call you.', '2016-08-15 00:18:11'), 
(7, '3om0e1abom4', 3, 4, 'Girl you\'re like a car accident, cause I just can\'t look away. 3=>4', '2016-08-15 00:42:57'), 
(8, '3om0e1abom4', 3, 4, 'Hola!! \r\n3=>4', '2016-08-15 00:43:34'), 
(9, '1omd0666om3', 3, 1, 'Sometext from 3=>1', '2016-08-15 13:53:54'), 
(10, '3om0e1abom4', 3, 4, 'More from 3->4', '2016-08-15 13:54:46'); 

Answer 1

讓我們試試這個（上fiddle）：

SELECT * 
FROM (SELECT * FROM user_msgs 
    WHERE from_id = 1 OR to_id = 1 
    ORDER BY id DESC 
) main 
GROUP BY from_id + to_id 
ORDER BY id DESC 

事情提GROUP BY from_id + to_id這是因爲總和使得它獨特的兩個人之間的每個對話：就像從1至3是一樣的，從3 1。無需額外的表，這使得它很難維持

UPDATE：

因爲有時GROUP ING在MySQL怪異的作品我創建的新方法，這一問題：在問題本身

SELECT 
    a.* 
FROM user_msgs a 
LEFT JOIN user_msgs b 
    ON ((b.`from_id` = a.`from_id` AND b.`to_id` = a.`to_id`) 
     OR (b.`from_id` = a.`to_id` AND b.`to_id` = a.`from_id`)) 
    AND a.`id` < b.`id` 
WHERE (a.from_id = 1 OR a.to_id = 1) 
    AND b.`id` IS NULL 
ORDER BY a.id DESC